Question: $h(t) = 4t+1$ $g(n) = -6n-2-2(h(n))$ $f(n) = 5n^{2}+3(h(n))$ $ f(h(0)) = {?} $
First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = (4)(0)+1$ $h(0) = 1$ Now we know that $h(0) = 1$ . Let's solve for $f(h(0))$ , which is $f(1)$ $f(1) = 5(1^{2})+3(h(1))$ To solve for the value of $f$ , we need to solve for the value of $h(1)$ $h(1) = (4)(1)+1$ $h(1) = 5$ That means $f(1) = 5(1^{2})+(3)(5)$ $f(1) = 20$